Find the LCD 1/(3x+15) , 1/(9x+45)

$\frac{1}{3 x + 15}$ , $\frac{1}{9 x + 45}$

Step 1

Simplify each polynomial.

Factor $3$ out of $3 x + 15$ .

Factor $3$ out of $3 x$ .

$\frac{1}{3 (x) + 15}, \frac{1}{9 x + 45}$

Factor $3$ out of $15$ .

$\frac{1}{3 x + 3 \cdot 5}, \frac{1}{9 x + 45}$

Factor $3$ out of $3 x + 3 \cdot 5$ .

$\frac{1}{3 (x + 5)}, \frac{1}{9 x + 45}$

Factor $9$ out of $9 x + 45$ .

Factor $9$ out of $9 x$ .

$\frac{1}{3 (x + 5)}, \frac{1}{9 (x) + 45}$

Factor $9$ out of $45$ .

$\frac{1}{3 (x + 5)}, \frac{1}{9 x + 9 \cdot 5}$

Factor $9$ out of $9 x + 9 \cdot 5$ .

$\frac{1}{3 (x + 5)}, \frac{1}{9 (x + 5)}$

Step 2

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$3 (x + 5), 9 (x + 5)$

Step 3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 4

Since $3$ has no factors besides $1$ and $3$ .

$3$ is a prime number

Step 5

$9$ has factors of $3$ and $3$ .

$3 \cdot 3$

Step 6

Multiply $3$ by $3$ .

$9$

Step 7

The factor for $x + 5$ is $x + 5$ itself.

$(x + 5) = x + 5$

$(x + 5)$ occurs $1$ time.

Step 8

The LCM of $x + 5, x + 5$ is the result of multiplying all factors the greatest number of times they occur in either term.

$x + 5$

Step 9

The Least Common Multiple $LCM$ of some numbers is the smallest number that the numbers are factors of.

$9 (x + 5)$